Problem 25 - Thomas Calculus

Question

Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$ f(x)=x^{2}-5, \quad L=11, \quad c=4, \quad \epsilon=1 $$

Step-by-Step Solution

Answer

The interval is \((3.877, 4.123)\), and \(\delta \approx 0.123\).

Step 1: Calculate Target Inequality

We need to find an interval around \(c\) for which \(|f(x) - L| < \epsilon\) holds. First, substitute the given function and values into the inequality: \[ |x^2 - 5 - 11| < 1 \Rightarrow |x^2 - 16| < 1. \] This is our target inequality.

Step 2: Solve the Target Inequality

Rewrite the inequality \(|x^2 - 16| < 1\) by considering the expression as a compound inequality: \[ -1 < x^2 - 16 < 1. \] Solving this inequality, we find \[ 15 < x^2 < 17. \] Next, take the square root of the entire inequality: \[ \sqrt{15} < x < \sqrt{17}. \] This is the interval for \(x\) around \(c = 4\).

Step 3: Determine the \\delta-Interval

Since we want \(0 < |x-c| < \delta\), where \(c = 4\), convert the inequalities for \(x\) such that they express \delta:\[ 4 - \sqrt{17} < \delta < \sqrt{17} - 4. \] Calculate these values explicitly:\[ \sqrt{17} \approx 4.123, \text{ hence } \delta = 0.123. \] Our open interval becomes \( (4 - 0.123, 4 + 0.123)\) or approximately \( (3.877, 4.123)\).

Key Concepts

Epsilon-delta definition of a limit

The concept of epsilon-delta is a formal way to define the limit of a function. When we say a function \( f(x) \) approaches a limit \( L \) as \( x \) approaches \( c \), our goal is to make the values of \( f(x) \) as close to \( L \) as we want, by keeping \( x \) close enough to \( c \). This closeness is measured with \( \epsilon \) and \( \delta \) values.

• \( \epsilon \) (epsilon) represents the distance we allow \( f(x) \) to deviate from its limit \( L \).
• \( \delta \) (delta) signifies how close \( x \) must be to \( c \) for the limit condition to hold true, i.e., \(0 < |x-c| < \delta\).

If for every \( \epsilon > 0\) there exists a \( \delta > 0\) such that whenever \(0 < |x-c| < \delta\), it follows that \(|f(x) - L| < \epsilon\), then \( \ \lim_{x \to c} f(x) = L \) by definition.
In our exercise, to find a suitable \( \delta \), we first ensure that \(|f(x) - L| < \epsilon \), obtaining our interval for \(x\). From this grounded understanding, we can determine the precise value of \( \delta \).

Inequalities in calculus

Inequalities are used throughout calculus to express a range of values within a specific scope. They can be used to bound a function near a certain point, helping us understand a function's limiting behavior.
In our exercise, the given inequality \(|x^2 - 16| < 1\) reflects how much \(x^2 - 16\) can vary to keep \( f(x) \) within an \(\epsilon\) distance from the limit \( L \), which is 11 in this case.
To solve the inequality \(|x^2 - 16| < 1\), break it into a compound inequality: \(-1 < x^2 - 16 < 1\). This helps isolate the variable \( x \), leading to the refined range \(15 < x^2 < 17\).
By taking the square root of these bounds, we discover the permissible values of \(x\) around \(c = 4\), specifically \(\sqrt{15} < x < \sqrt{17}\). This process reveals a span of values where the original inequality and therefore the limit holds.

Function behavior near a point

Analyzing function behavior near a specific point helps in understanding how a function approaches a limit. Around any given point \(c\), a function might behave differently, growing closer or farther with varying \(x\) values.
For the function \(f(x) = x^2 - 5\), calculated near \(c = 4\), we focus on the immediate vicinity of this point. By solving \(|x^2 - 16| < 1\), we see how \(f(x)\) acts in this region: it remains close to the target limit \(L = 11\) under the defined conditions.
The derived interval \( (\sqrt{15}, \sqrt{17}) \) reflects where \(x\) can lie to ensure \(|x^2 - 16| < 1\). This small area between \(3.877\) and \(4.123\) offers a glimpse into the subtleties of \(f(x)\) near \(c\), where it gently moves towards the limit value, consistently staying within \(\epsilon = 1\) from it.
These calculations provide insights into how a function smoothly transitions as it nears a particular point, aiding in the deep comprehension of limits and continuity.